Base | Representation |
---|---|
bin | 11100001100111001101111… |
… | …111001110110011001110011 |
3 | 121021011100010100212202202121 |
4 | 130030321233321312121303 |
5 | 112224114404402304112 |
6 | 1115443253224255111 |
7 | 35061004113251122 |
oct | 3414715771663163 |
9 | 537140110782677 |
10 | 124031943009907 |
11 | 3657a719360a19 |
12 | 11ab227a911497 |
13 | 5429214ccc8aa |
14 | 228b26c8b9ab9 |
15 | e5154eae1407 |
hex | 70ce6fe76673 |
124031943009907 has 2 divisors, whose sum is σ = 124031943009908. Its totient is φ = 124031943009906.
The previous prime is 124031943009893. The next prime is 124031943009929. The reversal of 124031943009907 is 709900349130421.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 124031943009907 - 215 = 124031942977139 is a prime.
It is not a weakly prime, because it can be changed into another prime (124031943009407) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 62015971504953 + 62015971504954.
It is an arithmetic number, because the mean of its divisors is an integer number (62015971504954).
Almost surely, 2124031943009907 is an apocalyptic number.
124031943009907 is a deficient number, since it is larger than the sum of its proper divisors (1).
124031943009907 is an equidigital number, since it uses as much as digits as its factorization.
124031943009907 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1469664, while the sum is 52.
The spelling of 124031943009907 in words is "one hundred twenty-four trillion, thirty-one billion, nine hundred forty-three million, nine thousand, nine hundred seven".
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